How to obtain $\int_0^{\infty} \frac{1}{\sigma^2 y + (N-1)\sigma_t^2} y^{\frac{N-1}{2}-1}e^{-\frac{y}{2}} dy$?

Question

Let $X_1, X_2 \cdots X_N$ be random variables, which follow a Gaussian distribution.

\begin{equation} X \sim N(\mu, \sigma^2) \end{equation} Let the parameters $\mu$ and $\sigma^2$ be unknown. The unbiased estimator of $\sigma^2$ is \begin{equation} U = \frac{1}{N-1}\left((X_1 -\bar{X})^2 \cdots (X_N -\bar{X})^2\right) \end{equation}

I'd like to calculate the mean value of $\tilde{F}=\frac{1}{U + \sigma_t^2}$ where $\sigma_t^2$ is a known parameter. \begin{equation} \begin{split} E\left[\tilde{F} \right]&= E\left[\frac{1}{U + \sigma_t^2}\right] \\ &= E\left[\frac{N-1}{\sigma^2 Y + (N-1)\sigma_t^2}\right] \quad (Y \sim \chi_{N-1}^2) \\ &= (N-1) \int_0^{\infty} \frac{1}{\sigma^2 y + (N-1)\sigma_t^2} \frac{y^{\frac{N-1}{2}-1}e^{-\frac{y}{2}}} {2^{\frac{N-1}{2}} \Gamma(\frac{N-1}{2})} dy \end{split} \end{equation} How can I calculate this value?

Answer 1

Consider \begin{equation} \begin{split} \int_0^{\infty} \frac{1}{\sigma^2 y + (N-1)\sigma_t^2} \frac{y^{\frac{N-1}{2}-1}e^{-\frac{y}{2}}} {2^{\frac{N-1}{2}} \Gamma(\frac{N-1}{2})} dy \end{split} \end{equation} and put $x = y+\frac{(N-1)\sigma_t^2}{\sigma^2}$.

After simplification we get $$A = const_1 \int_{const_2}^{\infty}x^{-1}(x-const_3)^{\frac{N-3}2}e^{-\frac{x}2}dx$$

If $\frac{N-3}2 = k \in \mathbb{Z}$ we can write binomial theorem for $(x-const_3)^{\frac{N-3}2}$ and put $x = 2 z$ and see that $A$ is a sum of incomplete gamma-functions.

If $\frac{N-3}2 \ne k \in \mathbb{Z}$ it looks like there's not good representation.